NUMERICAL PROBLEMS 

 F.Sc. Physics, 

Chapter # 12: ELECTROSTATICS 

12.1 Compare magnitudes of electrical and gravitational forces exerted on an object (mass = 10.0 g, charge = 20.0 µC) by an identical object that is placed 10.0 cm from the first. (G = 6.67 ×10 − 11 Nm2kg − 2) 

Given Data: 

Masses m1 = m2 = m = 10 g = 0.01 kg, 

Charges q1 = q2 = q = 20 µC = 20 × 10−6C, Distance r = 10 cm = 0.1 m 

To Determine:

 $\frac{F_{e}}{F_{g}} = ?$ 

Calculations: 

$\frac{F_{e}}{F_{g}} = \frac{\left( k\frac{q_{1}q_{2}}{r^{2}} \right)}{\left( G\frac{m_{1}m_{2}}{r^{2}} \right)} = \frac{kq_{1}q_{2}}{Gm_{1}m_{2}} = \frac{kq^{2}}{Gm^{2}} = \frac{9 \times 10^{9} \times \left( 20 \times 10^{- 6} \right)^{2}}{6.67 \times 10^{- 11} \times (0.01)^{2}} = 5.4 \times 10^{14}$ 12.2 Calculate vectorially the net electrostatic force on q as shown in the figure. Given Data: Charges q1 = 1 µC = 1 × 10−6C, q2 =  − 1 µC =  − 1 × 10−6C, q = 4 µC = 4 × 10−6C To Determine: Total Force on q F = ? Calculations: From Fig. $\tan\theta = \frac{0.8}{0.6} \Longrightarrow \theta = \tan^{- 1}\left( \frac{0.8}{0.6} \right) = 53{^\circ}$ Force Exerted by Charge q1 on q: $F_{1} = k\frac{qq_{1}}{r^{2}} = 9 \times 10^{9} \times \frac{4 \times 10^{- 6} \times 1 \times 10^{- 6}}{(1)^{2}} = 36 \times 10^{- 3}\ N$ Force Exerted by Charge q2 on q: $F_{2} = k\frac{qq_{2}}{r^{2}} = 9 \times 10^{9} \times \frac{4 \times 10^{- 6} \times 1 \times 10^{- 6}}{(1)^{2}} = 36 \times 10^{- 3}\ N$ Now Fx = F1x + F2x = F1cos θ + F2cos θ = 36 × 10−3 × cos 53∘ + 36 × 10−3 × cos (2π−53∘) = 0.043 N And Fy = F1y + F2y = F1sin θ + F2sin θ = 36 × 10−3 × sin 53∘ + 36 × 10−3 × sin (2π−53∘) = 0 N Magnitude of Resultant Force $F = \sqrt{F_{x}^{2} + F_{y}^{2}} = \sqrt{(0.043)^{2} + (0)^{2}} = 0.043\ N$ Direction of Resultant Force $\tan\theta = \frac{F_{y}}{F_{x}} \Longrightarrow \theta = \tan^{- 1}\left( \frac{0}{0.043} \right) = 0{^\circ}$ (Resultant is along x-axis) Resultant Force F⃗ = 0.043 î 12.3 A point charge q = − 8.0 × 10 − 8C is placed at the origin. Calculate electric field at a point 2.0 m from the origin on the z-axis. Given Data: Charge q = − 8.0 × 10−8C, Distance r = 2 m, Direction: z-axis r̂ = k̂ To Determine: Electric Field E = ? Calculations: $\overrightarrow{E} = k\frac{q}{r^{2}}\widehat{k} = 9 \times 10^{9} \times \frac{\left( - \ 8.0\ \times \ 10^{- 8} \right)}{(2)^{2}}\widehat{k} = \left( - 1.8 \times \ 10^{2}\ \widehat{k} \right)NC^{- 1}\ $ 12.4 Determine the electric field at the position $\overrightarrow{\mathbf{r}}\mathbf{\ = \ (4\ }\widehat{\mathbf{i\ }}\mathbf{+ \ 3}\widehat{\mathbf{j}}\mathbf{\ )m}$ caused by a point charge q = 5.0 × 10 − 6 C placed at origin. Given Data: Position Vector $\overrightarrow{r}\ = \ (4\ \widehat{i\ } + \ 3\widehat{j}\ )m$, Charge q = 5.0 × 10−6 C To Determine: Electric Field E⃗ = ? Calculations: $\overrightarrow{E} = k\frac{q}{r^{2}}\widehat{r}\ \ - - - - \ \ \ (1)$ $$Now\ r = |r| = \sqrt{4^{2} + 3^{2}} = 5,\ and\ \widehat{r} = \frac{\overrightarrow{r}}{|r|} = \frac{4\ \widehat{i\ } + \ 3\widehat{j}}{5}$$ Equation (1) becomes: $$\overrightarrow{E} = k\frac{q}{r^{2}}\widehat{r} = 9 \times 10^{9} \times \frac{\ 5.0\ \times \ 10^{- 6}}{(5)^{2}} \times \frac{4\ \widehat{i\ } + \ 3\widehat{j}}{5} = 360 \times \left( 4\ \widehat{i\ } + \ 3\widehat{j} \right) = 1440\ \widehat{i\ } + \ 1080\widehat{j}$$ 12.5 Two point charges, q1 =  − 1.0×10 − 6C and q2 = 4.0×10 − 6C, are separated by a distance of 3.0 m. Find and justify the zero-field location. Given Data: Charges q1 =  − 1.0 × 10−6C, q2 = 4.0 × 10−6C, Let Distance between Charges r = 3 m To Determine: Zero Field Location Calculations: Let P is zero field location, then at point P (distant x from q1): Electric Field due to q1= Electric Field due to q2 ⇒ E1 = E2 Let Distance of P from q1 = |AP| = r1 = x and Distance of P from q2 = |BP| = r2 = x + 3 Now consider $E_{1} = E_{2} \Longrightarrow k\frac{q_{1}}{r_{1}^{2}} = k\frac{q_{2}}{r_{2}^{2}} \Longrightarrow \frac{q_{1}}{r_{1}^{2}} = \frac{q_{2}}{r_{2}^{2}} \Longrightarrow \frac{1.0 \times 10^{- 6}}{x^{2}} = \frac{4.0 \times 10^{- 6}}{(x + 3)^{2}} \Longrightarrow \frac{1}{x^{2}} = \frac{4}{(x + 3)^{2}}$  ⇒ (x+3)2 = 4x2 ⇒ x2 + 6x + 9 = 4x2 ⇒ 3x2 − 6x − 9 = 0 ⇒ x2 − 2x − 3 = 0  ⇒ x2 − 3x + x − 3 = 0 ⇒ x(x−3) + 1(x−3) = 0 ⇒ (x−3)(x+1) = 0  ⇒ x − 3 = 0 or x + 1 = 0 ⇒ x = 3 or x =  − 1 (Not Possible) So the correct answer is x = 3 12.6 Find the electric field strength required to hold suspended a particle of mass 1.0×10 − 6 kg and charge 1.0 µC between two plates 10.0 cm apart. Given Data: Mass m = 1.0 × 10−6 kg, Charge q = 1.0 µC = 1.0 × 10−6 C , Distance between Plates d = 10 cm = 0.1 m To Determine: Electric Field Strength E = ? Calculations: For present case: Electrical Force = Gravitational Force ⇒ qE = mg $$\Longrightarrow E = \frac{mg}{q} = \frac{1.0 \times 10^{- 6} \times 9.8}{1.0 \times 10^{- 6}} = 9.8\ NC^{- 1}$$ 12.7 A particle having a charge of 20 electrons on it falls through a potential difference of 100 volts. Calculate the energy acquired by it in electron volts (eV). Given Data: Charge q = 20e, Potential Difference ΔV = 100 V To Determine: Energy Acquired  = ? Calculations: Energy Acquired  = qΔV = (20e)(100 V) = 2000 eV = 2 × 103eV 12.8 In Millikan’s experiment, oil droplets are introduced into the space between two flat horizontal plates, 5.00 mm apart. The plate voltage is adjusted to exactly 780 V so that the droplet is held stationary. The plate voltage is switched off and the selected droplet is observed to fall a measured distance of 1.50 mm in 11.2 s. Given that the density of the oil used is 900 kg m-3, and the viscosity of air at laboratory temperature is 1.0×10 − 6 Nm − 2s, calculate: a) The mass, and b) The charge on the droplet (Assuming g=9.8ms-2) Given Data: Distance between Plates d = 5 mm = 5 × 10−3m, Potential Difference V = 780 V Distance Covered Δs = 1.50 mm = 1.50 × 10−3m, Time t = 11.2 s, Density ρ = 900 kg m−3, Viscosity η = 1.0 × 10−6 Nm−2s To Determine: Mass of Droplet m = ?, Charge on Droplet q = ? Calculations: (a) Mass of Droplet $m = \frac{4}{3}\pi r^{3}\rho\ \ - - - - \ \ \ (1)$ Terminal Velocity $v_{t} = \frac{\mathrm{\Delta}s}{t} = \frac{{1.50 \times 10}^{- 3}}{11.2} = 0.1339 \times 10^{- 3}\ ms^{- 1}$ For a body moving with terminal velocity: $mg = 6\pi\eta rv \Longrightarrow \frac{4}{3}\pi r^{3}\rho g = 6\pi\eta rv_{t}\ \ \ \ \ \ \ \because By\ (1)$ $$\Longrightarrow r^{2} = \frac{9\eta v_{t}}{2g\rho} \Longrightarrow r = \sqrt{\frac{9\eta v_{t}}{2g\rho}} = \sqrt{\frac{9 \times 1.0 \times 10^{- 6} \times 0.1339 \times 10^{- 3}}{2 \times 9.8 \times 900}} = 1.1 \times 10^{- 6}\ m$$ Equation (1) becomes: $m = \frac{4}{3}\pi r^{3}\rho = \frac{4}{3} \times 3.14 \times \left( 1.1 \times 10^{- 6} \right)^{3} \times 900 = 5.018 \times 10^{- 15}kg$ (b) Charge $q = \frac{mgd}{V} = \frac{5.018 \times 10^{- 15} \times 9.8 \times 5 \times 10^{- 3}}{780} = 3.15 \times 10^{- 19}C$ 12.9 A proton placed in a uniform electric field of 5000 NC-1 directed to right is allowed to go a distance of 10.0 cm from A to B. Calculate Potential difference between the two points Work done by the field The change in P.E. of proton The change in K.E. of the proton Its velocity (mass of proton is 1.67 × 10 − 27 kg) Given Data: Electric Field E = 5000 NC−1, Distance covered d = 10 cm = 0.1 m, Charge on proton q = 1.6 × 10−19 C = 1e, Mass of proton m = 1.67 × 10−27kg To Determine: (a) Potential Difference V = ?, (b) Work Done W = ?, (c) Change in P.E. ΔU = ? (d) Change in K.E. ΔK = ?, (e) Velocity v = ? Calculations: (a) As $E = - \frac{\mathrm{\Delta}V}{d} \Longrightarrow V = - Ed = - 5000 \times 0.1 = - 500\ V$ (b) As $\mathrm{\Delta}V = \frac{W}{q} \Longrightarrow W = q\mathrm{\Delta}V = 1e \times 500\ V = - 500\ eV\ \ $ (c) ΔU =  − W =  − 500 eV ∵ − ve sign indicate decrease of P.E. (d) By work-energy principle: ΔK = W = 500 eV (e) As $\mathrm{\Delta}K = \frac{1}{2}mv^{2} \Longrightarrow v^{2} = \frac{2 \times \mathrm{\Delta}K}{m} \Longrightarrow v = \sqrt{\frac{2 \times \mathrm{\Delta}K}{m}} = \sqrt{\frac{2 \times 500\ eV}{1.67\ \times \ 10^{- 27}kg}} = \sqrt{\frac{2 \times 500 \times \ 1.6\ \times \ 10^{- 19}J}{1.67\ \times \ 10^{- 27}kg}}$  ⇒ v = 3.097 × 105 ms−1 12.10 Using zero reference point at infinity, determine the amount by which a point charge of 4.0×10 − 8 C alters the electric potential at a point 1.2 m away, when (a) Charge is positive (b) Charge is negative Given Data: Charge q = 4.0 × 10−8 C, Distance r = 1.2 m To Determine: (a) Electric Potential when charge is positive V+ = ?, (b) Electric Potential when charge is negative V− = ?, Calculations: (a) $V_{+} = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{r} = 9 \times 10^{9} \times \frac{4.0 \times 10^{- 8}}{1.2} = 300\ V$ (b) $V_{-} = \frac{1}{4\pi\varepsilon_{0}}\frac{- q}{r} = 9 \times 10^{9} \times \frac{\left( - 4.0 \times 10^{- 8} \right)}{1.2} = - 300\ V$ 12.11 In Bohr's atomic model of hydrogen atom, the electron is in an orbit around the nuclear proton at a distance of 5.29 ×10 − 11 m with a speed of 2.18 ×106 ms − 1. (e  = 1.60×10 − 19C, mass of electron = 9.10 ×10 − 31kg). Find The electric potential that a proton exerts at this distance Total energy of the atom in eV The ionization energy for the atom in eV Given Data: Distance r = 5.29  × 10−11 m , Speed v = 2.18  × 106 ms−1, Charge of Electron e  = 1.60 × 10−19C, Mass of Electron = 9.10  × 10−31kg To Determine: (a) Electric Potential due to proton V = ?, (b) Total Energy of atom E1 = ? (c) Ionization Energy of the atom  = ? Calculations: (a) $V = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{r} = 9 \times 10^{9} \times \frac{1.60 \times 10^{- 19}}{5.29\ \times 10^{- 11}} = 27.22\ V$ (b) From theory of atomic spectra, the energy of electron in nth orbit: $E_{n} = - \frac{ke^{2}}{2\ r_{n}}\ \ - - - - \ \ \ (1)$ For present case n = 1, so equation (1) takes the form: $E_{1} = - \frac{ke^{2}}{2\ r_{1}} = - \frac{9 \times 10^{9} \times \left( 1.60 \times 10^{- 19} \right)^{2}}{2 \times 5.29\ \times 10^{- 11}}$ $$= - 2.18 \times 10^{- 18}J = - \frac{2.18 \times 10^{- 18}}{1.60 \times 10^{- 19}} = - 13.6\ eV$$ (c) As electron possess 13.6 eV energy in the ground state of a H-atom. So, if we want to ionize such H-atom, we must supply 13.6 eV. Hence, the ionization energy of H-atom in ground state is 13.6 eV 12.12 The electronic flash attachment for a camera contains a capacitor for storing the energy used to produce the flash. In one such unit, the potential difference between the plates of a 750 µF capacitor is 330 V. Determine the energy that is used to produce the flash. Given Data: Potential Difference V = 330 V, Capacitance C = 750 pF = 750 × 10−6F To Determine: Energy U = ? Calculations: As $U = \frac{1}{2}CV^{2} = 0.5 \times 750 \times 10^{- 6} \times (330)^{2} = 40.83\ J\ \ $ 12.13 A capacitor has a capacitance of 2.5 ×10 − 8 F. In the charging process, electrons are removed from one plate and placed on the other one. When the potential difference between the plates is 450 V, how many electrons have been transferred?(e  = 1.60×10 − 19C) Given Data: Capacitance C = 2.5  × 10−8 F, Potential Difference V = 450 V, Charge q = e  = 1.60 × 10−19C To Determine: Total Number of Electrons Transferred n = ? Calculations: For a capacitor q = CV  −  − − (1), From Quantization of Charges q = ne  −  − − (2) Comparing (1) and (2):$\ ne = CV \Longrightarrow n = \frac{CV}{e} = \frac{2.5\ \times 10^{- 8} \times 450}{1.60 \times 10^{- 19}} = 7.03 \times 10^{13}$